A graph was plotted between molar conductivit | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) $HCl$ and $NaCl$ are strong electrolytes,so their molar conductivity decreases linearly with $\sqrt{C}$ according to the Debye-$H$ückel-Onsager eq

Vedclass · Accepted Answer

$I \ (HCl), II \ (NaCl), III \ (NH_4OH)$

Vedclass · Answer

(B) $HCl$ and $NaCl$ are strong electrolytes,so their molar conductivity decreases linearly with $\sqrt{C}$ according to the Debye-$H$ückel-Onsager equation: $\Lambda_m = \Lambda_m^0 - A\sqrt{C}$.$HCl$ has a higher molar conductivity than $NaCl$ due to the higher ionic mobility of $H^+$ ions compared to $Na^+$ ions.Therefore,curve $I$ corresponds to $HCl$ and curve $II$ corresponds to $NaCl$.$NH_4OH$ is a weak electrolyte,which shows a sharp increase in molar conductivity upon dilution (as $\sqrt{C}$ decreases),represented by curve $III$.

$A$ graph was plotted between molar conductivity of various electrolytes ($NaCl$,$HCl$ and $NH_4OH$) and $\sqrt{C}$ (in $mol \ L^{-1}$). Identify the correct assignment for the curves $I$,$II$,and $III$.

Similar Questions

Find out the ionisation constant of a weak acid $(HA)$ in terms of $\Lambda_m^o$ and $\Lambda_m^c$. (Given $\alpha$ cannot be ignored with respect to $1$)

The resistance of a $0.2 \ M$ solution of an electrolyte is $30 \ \Omega$ and conductivity is $1.2 \ S \ m^{-1}$. What is the value of the cell constant (in $cm^{-1}$)?

The molar conductivity for electrolytes $A$ and $B$ are plotted against $C^{1/2}$ as shown below. Electrolytes $A$ and $B$ respectively are:

Explain the difficulties encountered in the measurement of the resistance of ionic solutions and how to resolve them.

For a strong electrolyte,a plot of molar conductivity against (concentration)$^{1/2}$ is a straight line,with a negative slope. The correct unit for the slope is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module